A 50.00 mL sample of 0.500 M nitrous acid (HNO2) is titrated with 0.400 M NaOH.

Question

A 50.00 mL sample of 0.500 M nitrous acid (HNO2) is titrated with 0.400 M NaOH. Calculate the pH of the mixture after the addition of 35.0 mL of the NaOH solution.

Explanation / Answer

HNO2 + H2O---> HNO- + H3O+ Ka = [H3O+] [HNo-]/[HNO2] mole of HNO2 = molarity * volume = 0.5* 0.05 = 0.025 mole of NaOH added = 0.4 * 035 = 0.014 mole of HNO2 remained = 0.011 molarity = mole/total volume = 0.011/0.085 = 0.129 Now Ka for HNO2 = 4.3 * 10^-4 since OH- + HNO2 ---> HNO- + H2O [HNO-] = [OH-] added = 0.014/0.085 = 0.165 [HNO2] = 0.129 Ka = 4.3 * 10^-4 = [H+]0.165/0.129 => [H+] = 3.36 * 10^-4 hence pH = -log[H+] = -log[3.36*10^-4] = 3.47

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